A：签到。

#include
using namespace std;
#define ll long long
#define inf 1000000010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar(); return x*f; } int T,a[6],cnt[6]; signed main() { T=read(); while (T--) { for (int i=1;i<=5;i++) a[i]=read(); memset(cnt,0,sizeof(cnt)); for (int i=1;i<=5;i++) cnt[a[i]]++; int mx=0; for (int i=1;i<=5;i++) if (cnt[i]>cnt[mx]) mx=i;
cout<<mx<<endl;
}
return 0;
//NOTICE LONG LONG!!!!!
}

　　B：k是奇数时函数值均为1，k是偶数时每k+1个出现一个0。

#include
using namespace std;
#define ll long long
#define inf 1000000010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
ll read()
{
ll x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
return x*f;
}
int T;
ll l,r,k;
ll f(ll x)
{
return x-(x+1)/(k+1);
}
signed main()
{
T=read();
while (T--)
{
l=read(),r=read(),k=read();
if (k&1) cout<<r-l+1<<endl;
else cout<<f(r)-f(l-1)<<endl;
}
return 0;
//NOTICE LONG LONG!!!!!
}

　　C：暴力枚举横竖各切多少刀，将矩阵压成一行可以得到该情况下列的划分位置，压成一列可以得到该情况下行的划分位置，然后二维前缀和暴力验证即可。

#include
using namespace std;
#define ll long long
#define inf 1000000010
#define N 1010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
ll read()
{
ll x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar(); return x*f; } int T,n,m,k,tot,a[N][N],s[N],s2[N],s3[N][N],ans[N<<1],way[N<<1]; int calc(int x,int y,int l,int r) { return s3[y][r]-s3[x-1][r]-s3[y][l-1]+s3[x-1][l-1]; } signed main() { T=read(); while (T--) { n=read(),m=read(),k=read();tot=0; for (int i=1;i<=n;i++) for (int j=1;j<=m;j++) { char c=getc(); if (c=='0') a[i][j]=0;else a[i][j]=1,tot++; } for (int i=1;i<=n;i++) { s[i]=s[i-1]; for (int j=1;j<=m;j++) if (a[i][j]) s[i]++; } for (int i=1;i<=m;i++) { s2[i]=s2[i-1]; for (int j=1;j<=n;j++) if (a[j][i]) s2[i]++; } for (int i=1;i<=n;i++) for (int j=1;j<=m;j++) s3[i][j]=s3[i-1][j]+s3[i][j-1]-s3[i-1][j-1]+1-a[i][j]; tot=n*m-tot; bool isac=0; for (int i=1;i<=k;i++) ans[i]=2010; for (int i=0;i<=k;i++) if (tot%((i+1)*(k-i+1))==0) { int sum=tot/((i+1)*(k-i+1)); int last=0; int cnt=0; bool flag=1; for (int j=1;jsum*(k-i+1)) {flag=0;break;}
}
if (cntsum*(i+1)) {flag=0;break;}
}
}
if (cntans[j]) break;
}
}
}
}
if (!isac) printf("Impossible\n");
else
{
for (int i=1;i<k;i++) printf("%d ",ans[i]);
printf("%d\n",ans[k]);
}
}
return 0;
//NOTICE LONG LONG!!!!!
}

　　D：打表可知10k~10k+9范围内0~9各出现一次，于是只需要计算零散部分。

#include
using namespace std;
#define ll long long
#define inf 1000000010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
ll read()
{
ll x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar(); return x*f; } ll l,r,a[30],f[1000000]; int ff(ll n) { if (n<1000000) return f[n]; int cnt=0; while (n) a[++cnt]=n%10,n/=10; for (int j=1;j=1;j--) t=t*10+a[j]%10;
return ff(t);
}
ll calc(ll n)
{
int s=0;int w=ff(n/10*10);
for (ll i=n/10*10;i<=n;i++) s+=(w+(i-n/10*10))%10; return n/10*45+s; } signed main() { for (int i=0;i<=9;i++) f[i]=i; for (int i=10;i<=1000000;i++) { ll cnt=0,x=i; while (x) a[++cnt]=x%10,x/=10; for (int j=1;j=1;j--) t=t*10+a[j]%10;
f[i]=f[t];
}
int T=read();
while (T--)
{
l=read(),r=read();
cout<<calc(r)-calc(l-1)<<endl;
}
return 0;
//NOTICE LONG LONG!!!!!
}

　　先咕着。